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  1. Dynamic Array
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Dynamic Array

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54 Discussions

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  • michael_chen_0
     + 0 comments

    Python 3 solution

    def dynamicArray(n: int, queries: list[list[int]]) -> list[int]:
    arr: list[list[int]] = [[] for __ in range(n)]
    lastAnswer = 0
    answers: list[int] = []
    save_answer = answers.append
    for query in queries:
        subarray = arr[(query[1] ^ lastAnswer) % n]
        if query[0] == 1:
            subarray.append(query[2])
            continue
        save_answer(lastAnswer := subarray[query[2] % len(subarray)])
    return answers

  • outsider973
     + 0 comments

    My rust solution:

    fn dynamicArray(n: i64, queries: &[Vec<i64>]) -> Vec<i64> {
    let mut arr = vec![Vec::<i64>::new(); n as usize];
    let mut last_answer = 0;
    let mut answer = Vec::new();
    
    for query in queries {
        if query[0] == 1 {
            let idx = (query[1].bitxor(last_answer) % n) as usize;
            arr[idx].push(query[2]);
        } else {
            let idx = (query[1].bitxor(last_answer) % n) as usize;
            last_answer = arr[idx][(query[2] % arr[idx].len() as i64) as usize];
            answer.push(last_answer);
        }
    }
    answer
}

  • fadelabdelhamid1
     + 0 comments

    Where is the problem solving here? no problem to solve, no thinking, the problem is giving us the instructions to write ?

  • jcrijulshah
     + 0 comments
    def dynamicArray(n, queries):
    lastAnswer = 0
    arr = []
    answers = []
    for i in range(n):
        arr.append([])
    for query in queries:
        idx = (query[1]^ lastAnswer) % n
        if query[0] == 1:
            arr[idx].append(query[2])
        if query[0] == 2:
            lastAnswer = arr[idx][(query[2])% (len(arr[idx]))]
            answers.append(lastAnswer)
    return (answers)

  • shadowinc
     + 0 comments

    def ndimensional_list(dimensions):

    return [] if not dimensions else [ndimensional_list(dimensions[1:]) for _ in range(dimensions[0])]

    def dynamicArray(n, queries):

    arr = ndimensional_list([n])
lastAnswer = 0
n_arr = []
for i in queries:
    q = i[0]
    x = i[1]
    y = i[2]
    if q == 1:
        idx = (x ^ lastAnswer) % n
        arr[idx].append(y)
    if q == 2:
        idx = (x ^ lastAnswer) % n
        idy = y % len(arr[idx])
        lastAnswer = arr[idx][idy]
        n_arr.append(lastAnswer)
return n_arr